Simplify; express your answer in exponential form. Assume $k\neq 0, x\neq 0$. $\dfrac{{(k^{4})^{-5}}}{{(k^{5}x^{3})^{-3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${k^{4}}$ to the exponent ${-5}$ . Now ${4 \times -5 = -20}$ , so ${(k^{4})^{-5} = k^{-20}}$ In the denominator, we can use the distributive property of exponents. ${(k^{5}x^{3})^{-3} = (k^{5})^{-3}(x^{3})^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k^{4})^{-5}}}{{(k^{5}x^{3})^{-3}}} = \dfrac{{k^{-20}}}{{k^{-15}x^{-9}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-20}}}{{k^{-15}x^{-9}}} = \dfrac{{k^{-20}}}{{k^{-15}}} \cdot \dfrac{{1}}{{x^{-9}}} = k^{{-20} - {(-15)}} \cdot x^{- {(-9)}} = k^{-5}x^{9}$.